Question 3
Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
Let us write the oxidation number of each element involved in the given reaction as:
+2 -2 0 0 +1 -2
Cu O(s) + H2(g) → Cu(s) + H2 O(g)
Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Let us write the oxidation number of each element involved in the given reaction as:
+3 -2 +2 -2 0 +4 -2
Fe2 O3(s) + 3C O(g) → 2Fe(s) + 3C O2(g)
Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.
(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
Let us write the oxidation number of each element involved in the given reaction as:
+3 -1 +1 +3 -1 -3 +1 +1 -1 +3 -1
4B Cl3(g) + 3 Li Al H4(s) → 2B2 H6(g) + 3Li Cl(s) + 3 Al Cl3 (s)
In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.
(d) 2K(s) + F2(g) → 2K+F– (s)
Let us write the oxidation number of each element involved in the given reaction as:
0 0 +1 -1
2K(s) + F2(g) → 2K+ F– (s)
In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to – 1 in KF i.e., F2 is reduced to KF.
Hence, the above reaction is a redox reaction.
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
Let us write the oxidation number of each element involved in the given reaction as:
-3 +1 0 +2 -2 +1 -2
4 N H3(g) + 5 O2(g) → 4N O(g) + 6H2 O(g)
Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.
